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Math Help - Exponential model from 2 points

  1. #1
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    Exponential model from 2 points

    I'm given two points. The model has to look like y=ae^bx. I have to make a model that "fits the points shown in the table":


    I started to do this one based off of a example I found in the book, but I just get lost in the beginning.

    4=Ce^k(0)
    4/e^k(0)
    (1/3)=(4/e^k(0))(e^k7)

    That's where I am (if I'm even on the right track).
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  2. #2
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    Quote Originally Posted by BeSweeet View Post
    I'm given two points. The model has to look like y=ae^bx. I have to make a model that "fits the points shown in the table":


    I started to do this one based off of a example I found in the book, but I just get lost in the beginning.

    4=Ce^k(0)
    4/e^k(0)
    (1/3)=(4/e^k(0))(e^k7)

    That's where I am (if I'm even on the right track).
    Do you mean y = ae^{bx}

    If so:

    f(0) = a = 4

    f(7) = 4e^{7b} = \frac{1}{3}

    e^{7b} = \frac{1}{12}

    Solve for b.

    ---------------

    If y = ae^bx then you can't solve with two unknowns and only one usable equation
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  3. #3
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    After doing all that, I got (as my final answer):
    y=4e^(-3.55t)

    ... Which isn't right. What am I doing wrong?
    Last edited by mr fantastic; October 5th 2009 at 05:40 PM.
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  4. #4
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    Quote Originally Posted by BeSweeet View Post
    After doing all that, I got (as my final answer):
    y=4e^(-3.55t)

    ... Which isn't right. What am I doing wrong?
    Perhaps the exact value of b is required: b = - \frac{\ln 12}{7}.
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